Question

8*7^(2*x^2-x)-7*8^(2*x^2-x)=0

Answer 1

$8*7^{2x^2-x}-7*8^{2x^2-x}=0\\8*7^{2x^2-x}=7*8^{2x^2-x}\\(\frac{7}{8} )^{2x^2-x}=\frac{7}{8}\\2x^2-x=1\\ 2x^2-x-1=0\\D=1^2+4*2=9\\\sqrt{D} =3\\x_1=\frac{1-3}{2*2}=-\frac{1}{2} \\x_2=\frac{1+3}{2*2}=1$